Answer :
The refractive index of the liquid is equal to 43.53.
Here, since the light is emitted from 8 cm beneath the surface of the liquid and strikes 7.6 cm from the point directly above the source it forms a right-angled triangle from which we can calculate the angle of refraction or the critical angle X.
tan(X) = (opposite ÷ adjacent)
Here, the opposite side is equal to 7.6 cm and the adjacent side is equal to 8 cm.
tan(X) = (7.6 ÷ 8)
X = [tex]tan^{-1}[/tex](7.6 ÷ 8)
X = 43.53°
From snells Law,
Sin(X )= (n1 ÷ n2)
where n1 is the refractive index of the first medium and n2 is the refractive index of the second medium.
Here the first medium is air and the second medium is the liquid
Hence, the refractive index of air, n1 = 1.
To find the refractive index of the liquid, the equation can be rewritten as
n2 = n1 ÷ sin(X)
Substituting the values we get
n2 = 1 ÷ sin(43.53°)
n2 = 1.45.
Hence, the refractive index of the liquid is 1.45.
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